HELP!!!!!

Note that  m< BRP  and m<CRQ = 90       

And m<BRP + m<BPR  = 90

Therefore, m<BPR  = m<CRQ

And m< PBR = m <QCR = 90

Therefore, by AA congruency.....ΔPBR is similar to ΔRCQ

Then

PB/BR  = RC/CQ      

4/ BR  = RC/18       but RC  = 2BR  ...so

4/BR  = 2BR/ 18    cross-multiply

4*18  = 2BR^2

72  = 2BR^2      divide both sides by 2

36 = BR^2          take the square root of both sides

6  = BR      so.........CR  = 2BR  = 2*6   = 12

And by the Pythagorean Theorem

PR^2 =  (PB^2 + BR^2) =  4^2 + 6^2  = 16 + 36   = 52

Likewise

RQ^2  = (RC^2 + CQ^2) = 12^2 + 18^2  =  144 + 324 = 468

So, again by the Pythagorean Theorem:

PQ^2  = RQ^2 + PR^2

PQ^2 =  468 + 52

PQ^2  = 520      take the square root of both sides

PQ = sqrt(520)  = 2sqrt(130)  = about 22.8